Proof of Pythagorean Theorem using similar triangles:
Let ABC be a right triangle with altitude BD
Then by similar triangles,
ABD ≈ ACB ⇒ (AD/AB = AB/AC) ⇒ ( AB2 = AD*AC)
DBC ≈ BAC ⇒ (DC/CB = CB/AC) ⇒ ( CB2 = DC*AC)
Adding the expressions in the parentheses we have,
AB2 + BC2 = AD*AC + AC*DC
AB2 + BC2 = AC(AD + DC)
AB2 + BC2 = AC(AC)
AB2 + BC2 = AC2
BTW.....There is little historical evidence that Pythagoras ever had much to do with his own "Theorem" ......